>Hi,
>I have a list of numbers; I want a quick and efficient way to
>find the subset of these numbers which lie between two limits.
>However these numbers are determined within the program itself.
> if a >= b and a < c :
How about
newlist = filter(lambda x, l=<expr1>, h=<expr2>: l <= x < h, oldlist)
(This was the primary motivation for default arguments: to pass context
for lambda)